# Proof for a simplicial morphism

Sorry for the title but I can't come up with another one...

...

Suppose X and Y are simplicial sets and $f : X \rightarrow Y$ is a map such that:

1.) f maps dimensions the right way, that is for all $n \in \mathbb{N}^0$ we have $f_n := f : X_n \rightarrow Y_n$

2.) f commutes with all face maps in all dimensions, that is $f_{n-1} \circ d^n_k = d^n_k \circ f_n$ for all $n,k \in \mathbb{N}^0$ with $k \leq n$

3.) $f$ commutes with the degeneracy $s_n$ that is $f_{n+1}s_n=s_nf_n$ in any dimension $n \in \mathbb{N}^0$

(EDIT: So up to this point we only know about commutation with ONE degeneracy)


Is $f$ a simplicial morphism?

( In my particular situation X and Y are Kan in addition and moreover X is 2-coskeletal and Y is n-coskeletal for n>=2 )

If this only holds under additional assumptions, feel free to write just about that particular situation.

-
If I'm not missing something, 1) - 3) just say that f is a simplicial map (see Definition 1.2 in May's book on simplicial objects in algebraic topology - online: math.uchicago.edu/~may/PAPERS1965.html - or VIII.5 in MacLane's book on homology). What's the definition of a simplicial morphism ? – Ralph Nov 21 '11 at 21:35
A simplicial morphism is a simplicial map, so it is. – Fernando Muro Nov 21 '11 at 21:58
Mirco's condition (3) only mentions the last degeneracy s_n. So it's not obviously a simplicial map/morphism. – Tom Leinster Nov 21 '11 at 22:17